Projectile Motion. Chin- Sung Lin STEM GARAGE SCIENCE PHYSICS

Size: px

Start display at page:

Download "Projectile Motion. Chin- Sung Lin STEM GARAGE SCIENCE PHYSICS"

Linda Thornton
5 years ago
Views:

1 Projectile Motion Chin- Sung Lin

2 Introduction to Projectile Motion q What is Projectile Motion? q Trajectory of a Projectile q Calculation of Projectile Motion

3 Introduction to Projectile Motion q What is Projectile Motion? q Trajectory of a Projectile q Calculation of Projectile Motion

4 What is Projectile Motion?

5 Features of Projectile Motion? Thrown into the Air 2-D Motion Parabolic Path Affected by Gravity Determined by Initial Velocity

6 Definition: Projectile Motion Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

7 Definition: Projectile Motion Projectile motion refers to the 2-D motion of an object that is given an initial velocity and projected into the air at an angle. The only force acting upon the object is gravity. It follows a parabolic path determined by the effect of the initial velocity and gravitational acceleration.

8 Introduction to Projectile Motion q What is Projectile Motion? q Trajectory of a Projectile q Calculation of Projectile Motion

9 Trajectory (Path) of a Projectile

10 Trajectory (Path) of a Projectile

11 y v 0 x

12 y x

13 y x

14 y x

15 y q Velocity is changing and the motion is accelerated x

16 y q Velocity is changing and the motion is accelerated q The horizontal component of velocity (v x ) is constant x

17 y q Velocity is changing and the motion is accelerated q The horizontal component of velocity (v x ) is constant q Acceleration from the vertical component of velocity (v y ) x

18 y q Velocity is changing and the motion is accelerated q The horizontal component of velocity (v x ) is constant q Acceleration from the vertical component of velocity (v y ) g = -9.81m/s 2 q Acceleration due to gravity is constant, and downward q a = g = m/s 2 x

19 y q The horizontal and vertical motions are independent of each other q Both motions share the same time (t) g = -9.81m/s 2 x

20 y q The horizontal and vertical motions are independent of each other q Both motions share the same time (t) q The horizontal velocity...v x = v 0 q The horizontal distance... d x = v x t g = -9.81m/s 2 x

21 y q The horizontal and vertical motions are independent of each other q Both motions share the same time (t) q The horizontal velocity...v x = v 0 q The horizontal distance... d x = v x t g = -9.81m/s 2 q The vertical velocity v y = g t q The vertical distance d y = 1 / 2 g t2 x

22 Trajectory (Path) of a Projectile q The path of a projectile is the result of the simultaneous effect of the H & V components of its motion q H component constant velocity motion q V component accelerated downward motion q H & V motions are independent q H & V motions share the same time t q The projectile flight time t is determined by the V component of its motion

23 Horizontally Launched Projectile q H velocity is constant v x = v 0 q V velocity is changing v y = g t q H range: q V distance: d x = v 0 t d y = 1 / 2 g t2

24 Introduction to Projectile Motion q What is Projectile Motion? q Trajectory of a Projectile q Calculation of Projectile Motion

25 Calculation of Projectile Motion q Example: A projectile was fired with initial velocity v 0 horizontally from a cliff d meters above the ground. Calculate the horizontal range R of the projectile. v 0 d g t R

26 Strategies of Solving Projectile Problems q H & V motions can be calculated independently q H & V kinematics equations share the same variable t v 0 d g t R

27 Strategies of Solving Projectile Problems H motion: d x = v x t R = v 0 t V motion: d y = d = 1 / 2 g t2 t = sqrt(2d/g) So, R = v 0 t = v 0 * sqrt(2d/g) v 0 d g t R

28 Numerical Example of Projectile Motion H motion: d x = v x t R = v 0 t = 10 t V motion: d y = d = 1 / 2 g t2 t = sqrt(2 *19.62/9.81) = 2 s So, R = v 0 t = v 0 * sqrt(2d/g) = 10 * 2 = 20 m V 0 = 10 m/s m g = 9.81 m/s 2 t R

29 Exercise 1: Projectile Problem A projectile was fired with initial velocity 10 m/s horizontally from a cliff. If the horizontal range of the projectile is 20 m, calculate the height d of the cliff. V 0 = 10 m/s d g = 9.81 m/s 2 t 20 m

30 Exercise 1: Projectile Problem H motion: d x = v x t 20 = v 0 t = 10 t t = 2 s V motion: So, d y = d = 1 / 2 g t2 = 1 / 2 (9.81) 22 = m d = m V 0 = 10 m/s d g = 9.81 m/s 2 t 20 m

31 Exercise 2: Projectile Problem A projectile was fired horizontally from a cliff m above the ground. If the horizontal range of the projectile is 20 m, calculate the initial velocity v 0 of the projectile. V m g = 9.81 m/s 2 t 20 m

32 Exercise 2: Projectile Problem H motion: d x = v x t 20 = v 0 t V motion: d y = d = 1 / 2 g t2 t = sqrt(2 *19.62/9.81) = 2 s So, 20 = v 0 t = 2 v 0 v 0 = 20/2 = 10 m/s V m g = 9.81 m/s 2 t 20 m

33 Projectile Motion with Angles

34 Example: Projectile Problem H & V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the horizontal and vertical components of the initial velocity? g = 9.81 m/s 2 v y 20 m/s 60 o v x

35 Example: Projectile Problem H & V A projectile was fired from ground with 20. m/s initial velocity at 60-degree angle. What s the horizontal and vertical components of the initial velocity? V x = V cos θ = 20 m/s cos 60 o = 10 m/s V y = V sin θ = 20 m/s sin 60 o = m/s g = 9.81 m/s 2 v y 20 m/s 60 o v x

36 Example: Projectile Problem At the Top A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the velocity of the projectile at the top of its trajectory? v g = 9.81 m/s 2 v y 20 m/s 60 o t v x R

37 Example: Projectile Problem At the Top A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the velocity of the projectile at the top of its trajectory? V = V x = 10 m/s v g = 9.81 m/s 2 v y 20 m/s 60 o t v x R

38 Example: Projectile Problem Height A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the maximum height that the ball can reach? g = 9.81 m/s 2 v y 20 m/s 60 o v x h

39 Example: Projectile Problem Height A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the maximum height that the ball can reach? V f 2 = V i2 + 2gd d = m (0 m/s) 2 = (17.32 m/s) (-9.81 m/s 2 ) d g = 9.81 m/s 2 v y 20 m/s 60 o v x h

40 Example: Projectile Problem - Time A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How long will the ball travel before hitting the ground? g = 9.81 m/s 2 v y 20 m/s 60 o v x t

41 Example: Projectile Problem - Time A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How long will the ball travel before hitting the ground? V f = V i + gt t = 1.77 s 0 m/s = m/s + (-9.81 m/s2 ) t 1.77 s x 2 = 3.53 s g = 9.81 m/s 2 v y 20 m/s 60 o v x t

42 Example: Projectile Problem H Range A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How far will the ball reach horizontally? g = 9.81 m/s 2 v y 20 m/s 60 o v x R

43 Example: Projectile Problem H Range A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. How far will the ball reach horizontally? d x = V x t R = (10 m/s )(3.53 s) = 35.3 m g = 9.81 m/s 2 v y 20 m/s 60 o v x R

44 Example: Projectile Problem Final V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the final velocity of the projectile right before hitting the ground? g = 9.81 m/s 2 v y 20 m/s 60 o v x v fx v fy v SCIENCE f PHYSICS

45 Example: Projectile Problem Final V A projectile was fired from ground with 20 m/s initial velocity at 60-degree angle. What s the final velocity of the projectile right before hitting the ground? V fx = V x V fx = 10 m/s V fy = -V y V fx = m/s V f = sqrt (V fx 2 + V fy2 ) = 20 m/s θ = tan -1 (V fy /V fx ) = -60 o g = 9.81 m/s 2 v y 20 m/s 60 o v x v fx v fy v SCIENCE f PHYSICS

46 Example: Projectile Problem Max R A projectile was fired from ground with 20 m/s initial velocity. How can the projectile reach the maximum horizontal range? What s the maximum horizontal range it can reach? g = 9.81 m/s 2 20 m/s θ R

47 V & H Velocity Vectors of Projectile

48 Launch Angles of Projectile 75 o 60 o 45 o 30 o 15 o

49 Summary of Projectile Motion q What is Projectile Motion? q Trajectory of a Projectile q Calculation of Projectile Motion

50 The End

When we throw a ball :

PROJECTILE MOTION When we throw a ball : There is a constant velocity horizontal motion And there is an accelerated vertical motion These components act independently of each other PROJECTILE MOTION A